Exercise 1 You have a clinical study in which 10 patients will either get the standard treatment or...

SPH 247 Statistical Analysis of Laboratory Data 1

Exercise 1You have a clinical study in which 10

patients will either get the standard treatment or a new treatmentRandomize which 5 of the 10 get the new

treatment so that all possible combinations can result. Use Excel or R or another formal randomization method.

Instead, randomize so that in each pair of patients entered by date, one has the standard and one the new treatment (blocked randomization).

What are the advantages of each method?Why is randomization important?

April 2, 2013


Exercise 1 SolutionsFor the first situation, using Excel, you can (for

example) put the numbers 1–10 in column A, put five “Treatment” and five “Standard” in column B, put =rand() in column C. Then fix the numbers in column C and sort columns B and C by the random numbers.

For the second situation, you can put random numbers in column B and fix them, then in cell C2 (assuming a header row) put =IF(B2<B3,"A","B") and in cell C3 put =IF(B3<B2,"A","B"). Then copy the pair of cells to C4, C6, C8, and C10. Or use “Treatment” and “Standard” instead of “A” and “B”.

There are many other ways to do this.

April 2, 2013

SPH 247 Statistical Analysis of Laboratory Data 3April 2, 2013

Patient RandomTreatment

1 0.015712A

2 0.555195A

3 0.131912A

4 0.691906A

5 0.12621A

6 0.563621B

7 0.237196B

8 0.08503B

9 0.80344B

10 0.107165B

Patient RandomTreatment

1 0.015712A

2 0.08503B

3 0.107165B

4 0.12621A

5 0.131912A

6 0.237196B

7 0.555195A

8 0.563621B

9 0.691906A

10 0.80344B

Before Sorting After Sorting


Patient Random Treatment

1 0.2285A

2 0.470443B

3 0.939263B

4 0.106693A

5 0.41447A

6 0.504649B

7 0.491129A

8 0.628083B

9 0.796861B

10 0.167474A

IF(C2<C3,"A","B")

IF(C3<C2,"A","B")

IF(C4<C5,"A","B")

IF(C5<C4,"A","B")

IF(C6<C7,"A","B")

IF(C7<C6,"A","B")

IF(C8<C9,"A","B")

IF(C9<C8,"A","B")

IF(C10<C11,"A","B")

IF(C11<C10,"A","B")


CommentsRandomization is important to avoid bias,

approximately balance covariates, and provide a basis for analysis.

Blocked randomization can better control for time effects, but this particular version risks unblinding the next patient.

April 2, 2013


Exercise 2Analyze the testosterone levels from Rosner’s

endocrin data set in the same way as we did for the estradiol levels, using anova(lm())

Reanalyze estradiol and testosterone using lme() and verify that the results are the same. Here is the specificationlme(Testosterone ~ 1, random = ~1 | Subject,data=endocrin)

Repeat the analysis of the coop data for specimens 2 and 5 separately. Do the analysis both with the traditional ANOVA tables using lm() and with lme() and compare the results

April 2, 2013


> anova(lm(Testosterone ~ Subject,data=endocrin))Analysis of Variance Table

Response: Testosterone Df Sum Sq Mean Sq F value Pr(>F) Subject 4 571.66 142.915 22.041 0.002246 **Residuals 5 32.42 6.484 ---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Replication error variance is 6.484, so the standard deviation of replicates is 2.55 pg/mLThis compared to average levels across subjects from 15.55 to 36.25Estimated variance across subjects is (142.915 − 6.484)/2 = 68.2155Standard deviation across subjects is 8.26 pg/mLIf we average the replicates, we get five values, the standard deviation of which is 8.45.This contains both the subject and replication variance.


> lme(Estradiol ~ 1, random = ~1 | Subject,data=endocrin)Linear mixed-effects model fit by REML Data: endocrin Log-restricted-likelihood: -28.41785 Fixed: Estradiol ~ 1 (Intercept) 14.99

Random effects: Formula: ~1 | Subject (Intercept) ResidualStdDev: 8.434606 2.458254

Number of Observations: 10Number of Groups: 5 > lme(Testosterone ~ 1, random = ~1 | Subject,data=endocrin)Linear mixed-effects model fit by REML Data: endocrin Log-restricted-likelihood: -28.51958 Fixed: Testosterone ~ 1 (Intercept) 25.3

Random effects: Formula: ~1 | Subject (Intercept) ResidualStdDev: 8.259258 2.546372

Number of Observations: 10Number of Groups: 5


Specimen 2> anova(lm(Conc ~ Lab + Lab:Bat,data=coop,sub=Spc=="S2"))Analysis of Variance Table

Response: Conc Df Sum Sq Mean Sq F value Pr(>F) Lab 5 4.6640 0.93280 150.52 4.947e-14 ***Lab:Bat 12 1.3951 0.11626 18.76 9.484e-08 ***Residuals 18 0.1115 0.00620

April 2, 2013

The test for batch-in-lab is correct, but the test for lab is not—the denominator should beThe Lab:Bat MS, so F(5,12) = 0.93280/0.11626 = 8.0234 so p = 0.00157, still significantThe residual variance is 0.00620 (sd=0.0787). The estimated batch variance within labs is (0.11626 −0.00620)/2 = 0.05503 (sd = 0.2346).The estimated lab variance is (0.93280 − 0.11626)/(3×2) = 0.13609 (sd = 0.3689)


> lme(Conc ~1, random = ~1 | Lab/Bat,sub=Spc=="S2",data=coop)Linear mixed-effects model fit by REML Data: coop Subset: Spc == "S2" Log-restricted-likelihood: 7.383694 Fixed: Conc ~ 1 (Intercept) 0.3658333

Random effects: Formula: ~1 | Lab (Intercept)StdDev: 0.3689019

Formula: ~1 | Bat %in% Lab (Intercept) ResidualStdDev: 0.2345892 0.07872243

Number of Observations: 36Number of Groups: Lab Bat %in% Lab 6 18

April 2, 2013

The standard deviations for Lab, Batch, and Residual exactly match those from the lm() analysis


Specimen 5> anova(lm(Conc ~ Lab + Lab:Bat,data=coop,sub=Spc=="S5"))Analysis of Variance Table

Response: Conc Df Sum Sq Mean Sq F value Pr(>F) Lab 5 16.3277 3.2655 33.7688 1.539e-08 ***Lab:Bat 12 7.2049 0.6004 6.2088 0.0003072 ***Residuals 18 1.7406 0.0967 ---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

April 2, 2013

The test for batch-in-lab is correct, but the test for lab is not—the denominator should beThe Lab:Bat MS, so F(5,12) = 3.2655/0.6004 = 5.4389 so p = 0.00766, still significantThe residual variance is 0.0967 (sd=0.3110). The estimated batch variance within labs is (0.6004 −0.0967)/2 = 0.2519 (sd = 0.5018).The estimated lab variance is (3.2655 − 0.6004)/(3×2) = 0.4442 (sd = 0.6665)


> lme(Conc ~1, random = ~1 | Lab/Bat,sub=Spc=="S5",data=coop)Linear mixed-effects model fit by REML Data: coop Subset: Spc == "S5" Log-restricted-likelihood: -30.32728 Fixed: Conc ~ 1 (Intercept) 7.761389

Random effects: Formula: ~1 | Lab (Intercept)StdDev: 0.6664803

Formula: ~1 | Bat %in% Lab (Intercept) ResidualStdDev: 0.5018488 0.3109703

Number of Observations: 36Number of Groups: Lab Bat %in% Lab 6 18

April 2, 2013

The standard deviations for Lab, Batch, and Residual exactly match those from the lm() analysis

Exercise 1 You have a clinical study in which 10 patients will either get the standard treatment or...

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